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We are tasked with differentiating \( f(x) = \sqrt{x} \) using the first principles of differentiation, as given by the formula:

\[
f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}
\]

Here, \( f(x) = \sqrt{x} \), so let’s substitute that into the limit expression:

\[
f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}
\]

To simplify this expression, we multiply the numerator and denominator by the conjugate of the numerator:

\[
f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \cdot \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}
\]

This gives:

\[
f'(x) = \lim_{h \to 0} \frac{(\sqrt{x+h})^2 - (\sqrt{x})^2}{h(\sqrt{x+h} + \sqrt{x})}
\]

Simplifying the numerator:

\[
f'(x) = \lim_{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}
\]

Cancel the \( h \) in the numerator and denominator:

\[
f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}
\]

Now, as \( h \to 0 \), \( \sqrt{x+h} \to \sqrt{x} \). So the limit becomes:

\[
f'(x) = \frac{1}{2\sqrt{x}}
\]

Thus, the derivative of \( f(x) = \sqrt{x} \) from first principles is:

\[
f'(x) = \frac{1}{2\sqrt{x}}
\]

Would you like any further clarification or a deeper explanation of any steps?

Here are 5 related questions to explore:
1. How would you apply first principles to differentiate \( f(x) = x^2 \)?
2. What is the geometric interpretation of the derivative obtained for \( f(x) = \sqrt{x} \)?
3. How do higher-order derivatives work in first principles for more complex functions?
4. Can first principles be used to differentiate trigonometric functions, like \( \sin(x) \)?
5. How does the concept of limits apply when calculating derivatives?

**Tip:** When simplifying expressions using first principles, multiplying by the conjugate is a common strategy to handle square roots effectively.

Differentiate f(x) = √x from first principles given that f'(x) = lim (h → 0) [(f(x+h) - f(x))/h], h ≠ 0.

Learn how to differentiate the function f(x) = √x using the first principles of differentiation. This solution covers key concepts in calculus such as limits and the definition of derivatives. Perfect for students in Grades 11-12.

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